Recovering Force from a Survival Curve

A survival curve gives the probability that a life remains alive beyond each duration. The force of mortality describes the instantaneous rate of decrease in that probability, relative to the probability of having survived to that duration. When a survival curve is given directly, recovering the force is therefore a matter of differentiating the curve and scaling the derivative by the curve’s current value.

Throughout, let $T_x$ be the future lifetime of a life currently aged $x$, measured in years, and write $S_x(t)=P(T_x>t)$ for the probability of surviving more than $t$ years. Thus, $t$ is duration since age $x$, and the attained age at duration $t$ is $x+t$.

Definition: If $S_x(t)$ is differentiable and positive at duration $t$, the force of mortality at attained age $x+t$ is $$ \mu_{x+t}=-\frac{S'_x(t)}{S_x(t)}. $$ Here $S'_x(t)$ means the derivative with respect to duration $t$, holding the starting age $x$ fixed.

The formula is local: it uses the slope and the height of the survival curve at the same duration. The derivative $S'_x(t)$ is usually negative, because survival probabilities do not increase with duration. The negative sign therefore gives a nonnegative force when the curve is decreasing. Since $t$ is measured in years, the force is expressed per year.

The conditions matter. The ratio is defined only where $S_x(t)>0$, and the derivative must exist at the point where the force is requested. If the survival curve reaches zero, or has a corner at a duration, this formula does not give a finite, ordinary force at that point.

Why the Derivative-Over-Survival Formula Holds

The force at duration $t$ measures the instantaneous chance of death over a short additional interval, conditional on survival to $t$. For $h>0$, that conditional probability is the probability of surviving beyond $t$ but not beyond $t+h$, divided by the probability of surviving beyond $t$:

P(t<T_x\le t+h\mid T_x>t) =\frac{S_x(t)-S_x(t+h)}{S_x(t)}.

Divide by $h$, then take the limit as $h$ decreases to zero. When $S_x$ is differentiable and positive at $t$, the difference quotient converges to the negative derivative divided by the current survival probability:

\begin{aligned} \mu_{x+t} &=\lim_{h\downarrow 0} \frac{P(t<T_x\le t+h\mid T_x>t)}{h}\\ &=\lim_{h\downarrow 0} \frac{S_x(t)-S_x(t+h)}{hS_x(t)} =-\frac{S'_x(t)}{S_x(t)}. \end{aligned}

This also explains why the denominator is essential. The numerator’s rate of decline is an unconditional change in survival from the starting age $x$; dividing by $S_x(t)$ expresses that change relative to the lives still surviving at duration $t$.

Worked Applications

Before differentiating, identify the survival curve’s duration variable, its domain, and the point at which force is required. The derivative is with respect to duration, not with respect to a parameter or an unrelated age variable.

Worked Example: Force from an Exponential Survival Curve

A hypothetical group of lives aged 50 has a continuous future-lifetime model with survival function $S_{50}(t)=e^{-0.03t}$, where $t$ is measured in years and $t\ge 0$. Find the survival probability at five years and the force at attained age 55. State the force to four significant figures and the probability to four decimal places.

Method. Differentiate the specified survival curve, evaluate both the curve and derivative at $t=5$, and take their negative ratio. This avoids treating the survival probability itself as a force.

First evaluate survival: $$ S_{50}(5)=e^{-0.03(5)}=e^{-0.15}=0.86070798. $$

Differentiate with respect to $t$: $$ S'_{50}(t)=-0.03e^{-0.03t}, \qquad S'_{50}(5)=-0.03e^{-0.15}=-0.02582124. $$

The attained age is $50+5=55$, so $$ \mu_{55} =-\frac{-0.02582124}{0.86070798} =0.03000\text{ per year}. $$ The five-year survival probability is $0.8607$, and the force at age 55 is $0.03000$ per year. The check $0.03e^{-0.15}/e^{-0.15}=0.03$ confirms that the survival factor cancels.

Worked Example: A Polynomial Survival Curve

For a hypothetical life aged $x$, suppose the continuous future lifetime has survival function $S_x(t)=(1-t/20)^3$ for $0\le t\le20$, with $t$ in years. Calculate the force at durations 5 and 10 years, and compare them. Give the forces to four significant figures.

Method. Differentiate the polynomial and divide by the survival value at each duration. This avoids the common sign error of reporting the negative derivative itself as the force.

For $0\le t<20$, the survival value is positive. By the chain rule, $$ S'_x(t)=3(1-t/20)^2\left(-\frac{1}{20}\right) =-\frac{3}{20}(1-t/20)^2. $$

At $t=5$, $1-t/20=3/4$, so $$ S_x(5)=\left(\frac34\right)^3=\frac{27}{64}=0.421875, \qquad S'_x(5)=-\frac{3}{20}\left(\frac34\right)^2 =-\frac{27}{320}=-0.084375. $$ Therefore, $$ \mu_{x+5} =-\frac{-27/320}{27/64} =\frac{64}{320} =0.2000\text{ per year}. $$

At $t=10$, $1-t/20=1/2$, giving $$ S_x(10)=\left(\frac12\right)^3=\frac18=0.125, \qquad S'_x(10)=-\frac{3}{20}\left(\frac12\right)^2 =-\frac{3}{80}=-0.0375. $$ Thus, $$ \mu_{x+10} =-\frac{-3/80}{1/8} =\frac{24}{80} =0.3000\text{ per year}. $$ The force increases from $0.2000$ to $0.3000$ per year between these durations. Although the magnitude of the curve’s slope decreases here, the survival value in the denominator decreases faster, so the ratio increases.

Worked Example: Product Rule for a Polynomial Curve

A hypothetical continuous future-lifetime model has survival function $S_x(t)=(1-t/16)(1-t/24)$ for $0\le t\le16$, with $t$ in years. Find the force at durations 4 and 8 years, and determine which is larger. Report forces to four significant figures.

Method. Use the product rule to obtain the derivative, then evaluate the survival and derivative at both durations. Differentiating only one factor would omit part of the curve’s rate of change.

Differentiating both factors gives $$ S'_x(t) =-\frac{1}{16}(1-t/24)-\frac{1}{24}(1-t/16). $$

At $t=4$, the factors are $3/4$ and $5/6$: $$ S_x(4)=\frac34\cdot\frac56=\frac58, \qquad S'_x(4)=-\frac{1}{16}\cdot\frac56-\frac{1}{24}\cdot\frac34 =-\frac{5}{96}-\frac{1}{32} =-\frac{1}{12}. $$ Hence $$ \mu_{x+4} =-\frac{-1/12}{5/8} =\frac{8}{60} =\frac{2}{15} =0.1333\text{ per year}. $$

At $t=8$, the factors are $1/2$ and $2/3$: $$ S_x(8)=\frac12\cdot\frac23=\frac13, \qquad S'_x(8)=-\frac{1}{16}\cdot\frac23-\frac{1}{24}\cdot\frac12 =-\frac{1}{24}-\frac{1}{48} =-\frac{1}{16}. $$ Thus $$ \mu_{x+8} =-\frac{-1/16}{1/3} =\frac{3}{16} =0.1875\text{ per year}. $$ The force is larger at duration 8. Both results use the full derivative of the product and the survival value at the same duration.

Interpreting the Result and Avoiding Traps

A useful equivalent form, where $S_x(t)>0$, is $$ \mu_{x+t}=-\frac{d}{dt}\ln S_x(t). $$ It follows from the derivative rule for a logarithm: $d(\ln S_x(t))/dt=S'_x(t)/S_x(t)$. This form can make differentiation simpler when the survival curve is written as an exponential. It does not change the requirement that the curve be positive at the duration considered.

The ratio is unchanged if a curve is multiplied by a positive constant: for $c>0$, the curve $cS_x(t)$ has derivative $cS'_x(t)$, and $-cS'_x(t)/(cS_x(t))=-S'_x(t)/S_x(t)$. A properly defined conditional survival curve has $S_x(0)=1$, but this cancellation explains why the local ratio depends on relative decline rather than the curve’s overall scale.

Using the wrong sign. A decreasing survival curve has $S'_x(t)\le0$; the negative sign in the formula makes the force nonnegative.
Using the wrong denominator. Divide by $S_x(t)$, the probability of surviving to the duration in question, not by $S_x(0)$ unless that happens to be the same value.
Confusing duration and attained age. If the starting age is $x$, a calculation at duration $t$ gives $\mu_{x+t}$. The derivative is with respect to $t$.
Evaluating at a zero or a nondifferentiable point. At $S_x(t)=0$, the ratio is undefined; at a point where the derivative does not exist, this formula does not provide an ordinary force.

Key takeaway: At any duration where the survival curve is differentiable and positive, calculate the force by taking the negative derivative with respect to duration and dividing by the survival probability at that same duration. Check the sign, time unit, and attained age.

Check Your Understanding

For each hypothetical continuous future-lifetime model, calculate the requested force. Give forces to four significant figures and probabilities to four decimal places where requested.

A life aged $x$ has $S_x(t)=e^{-0.018t}$, with $t$ in years. Find $S_x(10)$ and $\mu_{x+10}$. (Answer: $S_x(10)=0.8353$; $\mu_{x+10}=0.01800$ per year.)
A life has $S_x(t)=(1-t/30)^2$ for $0\le t\le30$. Find the force at duration 6 years. (Answer: $0.08333$ per year.)
A survival curve is $S_x(t)=(1-t/10)^4$ for $0\le t\le10$. Find the force at duration 2 years. (Answer: $0.5000$ per year.)
A survival curve is $S_x(t)=(1-t/15)(1-t/30)$ for $0\le t\le15$. Find the force at duration 5 years. (Answer: $0.1400$ per year.)
For $S_x(t)=(1-t/12)^2$ on $0\le t\le12$, give the force for $0\le t<12$ and state whether the formula gives a finite force at $t=12$. (Answer: $\mu_{x+t}=2/(12-t)$ per year for $0\le t<12$; at $t=12$, the ratio is undefined because $S_x(12)=0$.)

Recovering Force from a Survival Curve

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